Count characters in a string whose ASCII values are prime

Given a string S. The task is to count and print the number of characters in the string whose ASCII values are prime.

Examples:

Input: S = “geeksforgeeks”
Output : 3
‘g’, ‘e’ and ‘k’ are the only characters whose ASCII values are prime i.e. 103, 101 and 107 respectively.

Input: S = “abcdefghijklmnopqrstuvwxyz”
Output: 6

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to generate all primes upto max ASCII value of character of string S using Sieve of Eratosthenes. Now, Iterate the string and get the ASCII value of each character. If the ASCII value is prime then increment the count. Finally, print the count.

Below is the implementation of the above approach:

C++

// C++ implementation of above approach 
#include <bits/stdc++.h> 
using namespace std; 
#define max_val 257 
  
// Function to find prime characters in the string 
int PrimeCharacters(string s) 
{ 
  
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    // THAN OR EQUAL TO max_val 
    // Create a Boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    vector<bool> prime(max_val + 1, true); 
  
    // 0 and 1 are not primes 
    prime[0] = false; 
    prime[1] = false; 
    for (int p = 2; p * p <= max_val; p++) { 
  
        // If prime[p] is not changed, then 
        // it is a prime 
        if (prime[p] == true) { 
  
            // Update all multiples of p 
            for (int i = p * 2; i <= max_val; i += p) 
                prime[i] = false; 
        } 
    } 
  
    int count = 0; 
  
    // Traverse all the characters 
    for (int i = 0; i < s.length(); ++i) { 
        if (prime[int(s[i])]) 
            count++; 
    } 
  
    return count; 
} 
  
// Driver program 
int main() 
{ 
    string S = "geeksforgeeks"; 
  
    // print required answer 
    cout << PrimeCharacters(S); 
  
    return 0; 
} 

Java

// Java implementation of above approach  
class Solution 
{ 
static final int max_val=257;  
  
// Function to find prime characters in the String  
static int PrimeCharacters(String s)  
{  
  
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS  
    // THAN OR EQUAL TO max_val  
    // Create a Boolean array "prime[0..n]". A  
    // value in prime[i] will finally be false  
    // if i is Not a prime, else true.  
    boolean prime[]= new boolean[max_val+1]; 
      
    //initilize the value 
    for(int i=0;i<=max_val;i++) 
    prime[i]=true; 
  
    // 0 and 1 are not primes  
    prime[0] = false;  
    prime[1] = false;  
    for (int p = 2; p * p <= max_val; p++) {  
  
        // If prime[p] is not changed, then  
        // it is a prime  
        if (prime[p] == true) {  
  
            // Update all multiples of p  
            for (int i = p * 2; i <= max_val; i += p)  
                prime[i] = false;  
        }  
    }  
  
    int count = 0;  
  
    // Traverse all the characters  
    for (int i = 0; i < s.length(); ++i) {  
        if (prime[(int)(s.charAt(i))])  
            count++;  
    }  
  
    return count;  
}  
  
// Driver program  
public static void main(String args[]) 
{  
    String S = "geeksforgeeks";  
  
    // print required answer  
    System.out.print( PrimeCharacters(S));  
  
}  
} 
//contributed by Arnab Kundu 

Python3

# Python3 implementation of above approach  
  
from math import sqrt 
  
max_val = 257 
  
# Function to find prime characters in the string  
def PrimeCharacters(s) : 
  
    # USE SIEVE TO FIND ALL PRIME NUMBERS LESS  
    # THAN OR EQUAL TO max_val  
    # Create a Boolean array "prime[0..n]". A  
    # value in prime[i] will finally be false  
    # if i is Not a prime, else true. 
    prime = [True] * (max_val + 1) 
  
    # 0 and 1 are not primes  
    prime[0] = False
    prime[1] = False 
    for p in range(2, int(sqrt(max_val)) + 1) : 
  
        # If prime[p] is not changed, then  
        # it is a prime  
        if (prime[p] == True) :  
  
            # Update all multiples of p  
            for i in range(2*p ,max_val + 1, p) : 
                prime[i] = False
  
    count = 0
  
    # Traverse all the characters  
    for i in range(len(s)) : 
        if (prime[ord(s[i])]) : 
            count += 1
              
    return count  
  
# Driver program  
if __name__ == "__main__" :  
  
    S = "geeksforgeeks";  
  
    # print required answer  
    print(PrimeCharacters(S))  
  
# This code is contributed by Ryuga 

C#

// C# implementation of above approach  
using System; 
class GFG{ 
      
static readonly int max_val = 257;  
  
// Function to find prime characters in the String  
static int PrimeCharacters(String s)  
{  
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS  
    // THAN OR EQUAL TO max_val  
    // Create a Boolean array "prime[0..n]". A  
    // value in prime[i] will finally be false  
    // if i is Not a prime, else true.  
    bool []prime = new bool[max_val + 1];  
  
    //initilize the value  
    for(int i = 0; i <= max_val; i++)  
    prime[i] = true;  
  
    // 0 and 1 are not primes  
    prime[0] = false;  
    prime[1] = false;  
    for (int p = 2; p * p <= max_val; p++)  
    {  
        // If prime[p] is not changed, then  
        // it is a prime  
        if (prime[p] == true)  
        {  
            // Update all multiples of p  
            for (int i = p * 2; i <= max_val; i += p)  
            prime[i] = false;  
              
        }  
          
    }  
      
    int count = 0;  
      
    // Traverse all the characters  
    for (int i = 0; i < s.Length; ++i)  
    {  
        if (prime[(int)(s[i])])  
        count++;  
          
    }  
    return count;  
      
}  
  
// Driver Code 
public static void Main()  
{  
    String S = "geeksforgeeks";  
      
    // print required answer  
    Console.Write( PrimeCharacters(S));  
      
}  
}  
  
// This code is contributed by PrinciRaj1992 

Output:

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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